Two wires of the same material (Young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight W is suspended from the combination as shown in the figure. The elastic potential energy in the system is

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$\frac{3 W^{2} L}{4 π R^{2} Y}$

$\frac{3 W^{2} L}{8 π R^{2} Y}$

$\frac{5 W^{2} L}{8 π R^{2} Y}$

$\frac{W^{2} L}{π R^{2} Y}$

answer is C.

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Detailed Solution

$k_{1} = \frac{Y π {(2 R)}^{2}}{L}, k_{2} = \frac{Y π {(R)}^{2}}{L}$
Hence, equivalent force constant k
$\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}}$
[As two wires are joined in series]
= $\frac{L}{4 Y π R^{2}} + \frac{L}{Y π R^{2}}$ since $k_{1} x_{1} = k_{2} x_{2} = W$
Hence, elastic potential energy of the system.
$U = \frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{2} x_{2}^{2} \frac{1}{2} k_{1} {(\frac{W}{k_{1}})}^{2} + k_{2} \frac{1}{2} {(\frac{W}{k_{2}})}^{2} \frac{1}{2} W^{2} [\frac{1}{k_{1}} + \frac{1}{k_{2}}] \frac{1}{2} W^{2} [\frac{5 L}{4 Y π R^{2}}] = \frac{5 W^{2} L}{8 π Y R^{2}}$
Hence the correct answer is option(c).